Changing the variable inside an R formula

I recently encountered a situation where I wanted to run several linear models, but where the response variables would depend on previous steps in the data analysis pipeline. Let me illustrate using the mtcars dataset:

data(mtcars)
head(mtcars)
#>                    mpg cyl disp  hp drat    wt  qsec vs am gear carb
#> Mazda RX4         21.0   6  160 110 3.90 2.620 16.46  0  1    4    4
#> Mazda RX4 Wag     21.0   6  160 110 3.90 2.875 17.02  0  1    4    4
#> Datsun 710        22.8   4  108  93 3.85 2.320 18.61  1  1    4    1
#> Hornet 4 Drive    21.4   6  258 110 3.08 3.215 19.44  1  0    3    1
#> Hornet Sportabout 18.7   8  360 175 3.15 3.440 17.02  0  0    3    2
#> Valiant           18.1   6  225 105 2.76 3.460 20.22  1  0    3    1

Let’s say I wanted to fit a linear model of mpg vs. hp and get the coefficients. This is easy:

lm(mpg ~ hp, data = mtcars)$coefficients
#> (Intercept)          hp 
#> 30.09886054 -0.06822828 

But what if I wanted to fit a linear model of y vs. hp, where y is a response variable that I won’t know until runtime? Or what if I want to fit 3 linear models: each of mpg, disp, drat vs. hp? Or what if I want to fit 300 such models? There has to be a way to do this programmatically.

It turns out that there are at least 4 different ways to achieve this in R. For all these methods, let’s assume that the responses we want to fit models for are in a character vector:

response_list <- c("mpg", "disp", "drat")

Here are the 4 ways I know (in decreasing order of preference):

1. as.formula()

as.formula() converts a string to a formula object. Hence, we can programmatically create the formula we want as a string, then pass that string to as.formula():

for (y in response_list) {
    lmfit <- lm(as.formula(paste(y, "~ hp")), data = mtcars)
    print(lmfit$coefficients)
}
#> (Intercept)          hp 
#> 30.09886054 -0.06822828 
#> (Intercept)          hp 
#>    20.99248     1.42977 
#> (Intercept)          hp 
#>  4.10990867 -0.00349959 

2. Don’t specify the data option

Passing the data = mtcars option to lm() gives us more succinct and readable code. However, lm() also accepts the response vector and data matrix themselves:

for (y in response_list) {
    lmfit <- lm(mtcars[[y]] ~ mtcars$hp)
    print(lmfit$coefficients)
}
#> (Intercept)          hp 
#> 30.09886054 -0.06822828 
#> (Intercept)          hp 
#>    20.99248     1.42977 
#> (Intercept)          hp 
#>  4.10990867 -0.00349959 

Edit: Commenter Tommaso Gennari shared a really nice solution that makes use of the fact that when you give lm() just a data frame, the first column is used as a dependent variable and the remaining columns are treated as independent variables:

for (y in response_list) {
    lmfit <- lm(mtcars[, c(y, "hp")])
    print(lmfit$coefficients)
}
#> (Intercept)          hp 
#> 30.09886054 -0.06822828 
#> (Intercept)          hp 
#>    20.99248     1.42977 
#> (Intercept)          hp 
#>  4.10990867 -0.00349959 

3. get()

get() searches for an R object by name and returns that object if it exists.

for (y in response_list) {
    lmfit <- lm(get(y) ~ hp, data = mtcars)
    print(lmfit$coefficients)
}
#> (Intercept)          hp 
#> 30.09886054 -0.06822828 
#> (Intercept)          hp 
#>    20.99248     1.42977 
#> (Intercept)          hp 
#>  4.10990867 -0.00349959 

4. eval(parse())

This one is a little complicated. parse() returns the parsed but unevaluated expressions, while eval() evaluates those expressions (in a specified environment).

for (y in response_list) {
    lmfit <- lm(eval(parse(text = y)) ~ hp, data = mtcars) print(lmfit$coefficients) } #> (Intercept)          hp 
#> 30.09886054 -0.06822828 
#> (Intercept)          hp 
#>    20.99248     1.42977 
#> (Intercept)          hp 
#>  4.10990867 -0.00349959 

 

Of course, for any of these methods, we could replace the outer loop with apply() or purrr::map().

References:

  1. johnramey. Converting a String to a Variable Name On-The-Fly and Vice-versa in R.
Advertisements

3 thoughts on “Changing the variable inside an R formula

  1. Actually you could simplify even more way n.2 using: lm(mtcars[,c(y,”hp”)])
    (I have not tested this expression and there might be a detail I am not foreseeing; however my point is that when you feed into lm just a data frame, the first column is used as a dependent variable, and all the remaining as independent) – hope this helps!

    Like

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s